Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

So let $D\subseteq \mathbb{C}^n$ be a bounded connected open set with a transitive action of its group of biholomorphisms (which we denote by $Hol(D)$). Note that I'm not assuming that $D$ is symmetric. We thus have that $D$ is "homeomorphic" to $Hol(D)/K$ where $K=Stab(d_0)$ for some $d_0\in D$.

In the special case where $Hol(D)$ is a real Lie group and that $K$ is a maximal compact of $Hol(D)$ then by a theorem of Elie Cartan we have that $Hol(D)/K$ is homeomoprphic to $\mathbb{R}^m$ and thus contractible.

Under my assumptions:

(1) Is $Hol(D)$ always a Lie group?

(2) Is $K$ always a maximal compact?

(3) In general is $D$ always contractible (or simply connected)?

share|improve this question
3  
Dear Hugo -- re "is Hol(D) always a Lie group": no, take $D=\mathbb{C}^2$; it has automorphisms of the form $(x,y)\mapsto (x,y+f(x))$ where $f$ is any holomorphic function $\mathbb{C}\to\mathbb{C}$. –  algori Sep 8 '11 at 22:40
3  
Maybe I am missing something: isn't $D=\mathbb{C}^\times$ an example where $D$ is not contractible? –  M P Sep 8 '11 at 22:40
    
Yes you are right $\mathbb{C}^{\times}$ is a counter-example, so I'll redit my question –  Hugo Chapdelaine Sep 8 '11 at 23:09
    
I forgot put that $D$ was bounded –  Hugo Chapdelaine Sep 8 '11 at 23:21

2 Answers 2

up vote 2 down vote accepted

Re question 3: a bounded homogeneous domain is biholomorphic to a Siegel domain, which is contractible. See e.g. Siegel domain and references therein (those references probably answer question 2 as well). Another useful link is Homogeneous bounded domain.

upd: Another Google search gave the following references:

"Homogeneous Bounded Domains and Siegel Domains" by Soji Kaneyuki, Springer LNM 241.

"Theory of complex homogeneous bounded domains" by Yichao Xu, Mathematics and its applications 569.

share|improve this answer
    
Thanks a lot @algori for the reference –  Hugo Chapdelaine Sep 9 '11 at 1:02
2  
I fixed the links. –  George Lowther Dec 2 '11 at 1:28

It is a theorem of H. Cartan that $Hol(D)$ for any bounded such $D\subset \mathbb C^n$ is a finite dimensional real Lie group. See for example chapter 9 of "Several complex variables" by R. Narasimhan.

share|improve this answer
    
thanks @Gjergji –  Hugo Chapdelaine Sep 9 '11 at 1:04

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.